Calculate the ground state `Q` value of the induced fission reaction in the equation
`n+._(92)^(235) Urarr._(92)^(236)Uastrarr._(40)^(99)Zr+._(52)^(134)Te+2n`
If the neutron is thermal. A thermal neutron is in thermal equilibrium with its environment, it has an average kinetic energy given by `(3//2) kT`. Given :
`m(n) =1.0087 am u, M(.^(235)U)=235.0439) am u`,
`M(.^(99)Zr)=98.916 am u, M(.^(134)Te)=133.9115 am u`.

Answer

`184.84 MeV`

Explanation

kinetic energy of a thermal neutron can be neglected, even for a temperature of `10^(6)K`, the thermal energy is only `130 eV` The `Q` value of the above reaction is given by the equation
`Q =Delta m xx 931.2 MeV`
Here, `Delta m` is the mass defect of the reaction, given by
`Delta m=[M(.^(235)U) +m(n)] -[M(.^(99)Zr) +M(.^(134)Te) +3 m(n) ]`
`=[235.0439 - 98.9165 -133 .9115`
`-2(1.0087)]a.m.u.`
`=0.1985 a.m.u`.
Thus, energy released is `Q=0.1985 xx 931.2`. Here, `Delta m` is the mass defect of the reaction, given by
`Delta m=[M(.^(235)U) +m(n)] -[M(.^(99)Zr) +M(.^(134)Te) +3 m(n) ]`
Thusm energy released is
`Q=0.1985 xx 931.2 =184.84 MeV`
Even with a thermal neutron of negligible kinetic energy, a tremendous amount of energy is released.